Free FE Mechanical Practice Questions

Sample Practice Questions

Try these 10 sample questions from the FE Mechanical practice bank. Each includes the answer and explanation. Continue with the interactive practice above to review the full question set.

1. 1What is the slope of the straight line through (2, 5) and (6, 17)?

   1. A. 1/3
   2. B. 2
   3. C. 3
   4. D. 6

   Answer: C

   Explanation: The handbook states, "For a straight line, slope is equal to rise over run". Thus m = (17 − 5)/(6 − 2) = 3. Reversing the ratio gives 1/3; using only one point or an incorrect coordinate difference does not give the line slope.

   Reference: DOE Fundamentals Handbook: Mathematics, Volume 1 — Module MA-02, Slopes, p. 85

2. 2A continuous normal distribution is centered at 50 mm. What is the probability that a randomly selected value is below 50 mm?

   1. A. 0.25
   2. B. 0.50
   3. C. 0.68
   4. D. 0.95

   Answer: B

   Explanation: NIST states, "The normal distribution is a symmetric distribution with well-behaved tails and a single peak at the center of the distribution." Symmetry places half of the total probability below the center. The other probabilities confuse one side of the distribution with percentages associated with particular intervals.

   Reference: NIST/SEMATECH e-Handbook: Measures of Scale — Section 1.3.5.6, Normal Distribution example

3. 3What is the present worth of a single $12,100 receipt at the end of year 2 when the effective annual discount rate is 10%?

   1. A. $9,801
   2. B. $10,000
   3. C. $11,000
   4. D. $14,641

   Answer: B

   Explanation: NIST states, "Single present value (SPV) factor is used to calculate the present value (PV) of a future cash amount (Ft) occurring at the end of year t given discount rate d." Its displayed discount formula gives P = F/(1 + d)^t = 12,100/1.10² = $10,000. Discounting one year, compounding instead, or over-discounting gives the other values.

   Reference: NIST Handbook 135: Life Cycle Costing Manual — Section 3.2, PV formula for one-time amounts, p. 23; Section 17.2.1.1

4. 4A steady flow transfers 12 C of charge through a conductor in 4 s. What is the current magnitude?

   1. A. 0.33 A
   2. B. 48 A
   3. C. 16 A
   4. D. 3 A

   Answer: D

   Explanation: DOE states, "One ampere of current is defined as the movement of one coulomb of charge past any given point of a conductor during one second of time." Thus I = Q/t = 12/4 = 3 A. Reversing the ratio or multiplying charge by time does not give current.

   Reference: DOE Fundamentals Handbook: Electrical Science, Volume 1 — Module ES-01, Electrical Terminology, Current

5. 5A particle moves along a straight line with initial velocity 3 m/s and constant acceleration 2 m/s² in the same direction. What is its velocity after 4 s?

   1. A. 5 m/s
   2. B. 8 m/s
   3. C. 11 m/s
   4. D. 14 m/s

   Answer: C

   Explanation: DOE states, "Acceleration is a measure of the change in velocity or velocity per unit time". For constant acceleration, v = v0 + at = 3 + 2(4) = 11 m/s. The alternatives omit the initial velocity, omit elapsed time, or combine the quantities incorrectly.

   Reference: DOE Fundamentals Handbook: Classical Physics — Module CP-01, Unit Systems, derived quantities, Acceleration

6. 6What characterizes the principal coordinate directions of the stress state at a point?

   1. A. All normal stress components vanish.
   2. B. All normal stress components become equal.
   3. C. All shear stress components become maximum.
   4. D. All shear stress components vanish.

   Answer: D

   Explanation: SOLIDWORKS explains that at the principal orientation "all shear stresses vanish". The remaining components are principal normal stresses. These normal stresses need not be zero or equal; principal planes are not the maximum-shear planes.

   Reference: SOLIDWORKS Design Help: Principal Stresses Definition — Principal Stresses Definition, paragraph describing orientation X’Y’Z’

7. 7A liquid has a density of 800 kg/m³. What is its specific volume?

   1. A. 0.000800 m³/kg
   2. B. 1.25 m³/kg
   3. C. 0.00125 m³/kg
   4. D. 800 m³/kg

   Answer: C

   Explanation: The handbook states, "Density and specific volume are the inverse of one another." Therefore v = 1/ρ = 1/800 = 0.00125 m³/kg. Density itself and density with a decimal shift are not specific volume.

   Reference: DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow, Volume 3 — Module 3, Continuity Equation, page 2, Properties of Fluids

8. 8Air is modeled as an ideal gas with R = 0.287 kJ/(kg·K). At an absolute pressure of 200 kPa and a temperature of 300 K, what is its specific volume?

   1. A. 0.215 m³/kg
   2. B. 2.32 m³/kg
   3. C. 4.31 m³/kg
   4. D. 0.431 m³/kg

   Answer: D

   Explanation: DOE gives the ideal-gas equation as "Pv = RT" and specifies absolute pressure and temperature. Thus v = RT/P = (0.287)(300)/200 = 0.4305 m³/kg. Using gauge pressure or inverting the result would not produce the requested specific volume.

   Reference: DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow, Volume 1 — Module 1, Compression Processes, page 98, Equation (1-43)

9. 9Which temperature sensor uses a fine metallic sensing element whose electrical resistance changes with temperature?

   1. A. Thermocouple
   2. B. Resistance temperature detector
   3. C. Bimetallic thermometer
   4. D. Optical pyrometer

   Answer: B

   Explanation: DOE states that an RTD produces "a proportional change in resistance" when its temperature changes. A resistance temperature detector infers temperature from that electrical resistance. A thermocouple uses thermoelectric voltage, a bimetallic thermometer uses differential expansion, and a pyrometer detects radiation.

   Reference: DOE Fundamentals Handbook: Instrumentation and Control, Volume 1 — Module 1, Resistance Temperature Detectors, page 4, Summary

10. 10A shaft shoulder has an elastic stress concentration factor Kt = 2.4 for the specified loading and nominal-stress definition. If its nominal stress is 50 MPa and the response remains elastic, what is the predicted local peak stress?

    1. A. 20.8 MPa
    2. B. 70 MPa
    3. C. 120 MPa
    4. D. 240 MPa

    Answer: C

    Explanation: COMSOL uses the phrase "product of" and identifies the factors as "K_{\mathrm t} and a suitably selected nominal stress." For the specified consistent nominal-stress definition, σpeak = Ktσnom = 2.4(50) = 120 MPa. Dividing by Kt or adding it to the stress misapplies this elastic concentration factor.

    Reference: How Should I Evaluate Singular Stress Fields? — A Closer Look At Singular Stress Fields, stress-concentration discussion